Proof that limit $\frac{\ln(x)}x$ for $x\to\infty$ is $0$.

Question

Consider this proof of $\lim_{x\to{+\infty}} \frac{\ln(x)}{x} =0 $.

Using the simple substitution $x\mapsto x^n$ we have that $\forall n\in\mathbf{N}$ $$\lim_{x\to{+\infty}} \frac{\ln(x)}{x} = \lim_{x\to{+\infty}} \frac{n\cdot \ln(x)}{x^n}$$ and so we can write

$$\lim_{x\to{+\infty}} \frac{\ln(x)}{x} = \lim_{n\to{+\infty}} \lim_{x\to{+\infty}} \frac{n\cdot \ln(x)}{x^n}= \lim_{x\to{+\infty}} \lim_{n\to{+\infty}} \frac{n\cdot \ln(x)}{x^n}.$$

Now, the inner limit is 0 for all $x>1$ so it leaves $$\lim_{x\to{+\infty}} \frac{\ln(x)}{x^n} = \lim_{x\to{+\infty}} 0 = 0.$$

Is this proof correct?

Answer 1

Your proof corresponds to use an exponential change of variable as $x=e^y$ with $y \to \infty$ that is

$$\lim_{x\to{+\infty}} \frac{\ln(x)}{x}=\lim_{y\to{+\infty}} \frac{y}{e^y}$$

and you are assuming that the latter, in the form $\frac n{x^n}$ in your case, is true but we also need to prove that fact.

To conlcude we can use that eventually $e^y\ge y^2$ which can be proved easily by induction for $y\in \mathbb N$ and then extended to reals.

Answer 2

Personally, when two functions $g(x)$ and $f(x)$ both tend to infinity for $x→+∞$ but with different "speeds", I consider the hierarchy of infinites. The function $g(x)=x$ to the denominator dominates the function $f(x)=\ln x$ to the numerator because it goes faster than the $\ln x$ as you can see with this graph created with Desmos:

It's practically as if I had a constant at the numerator and an infinitely great function at the denominator. Therefore:

$$\lim_{x\to 0}\frac{\ln(x)}x=0$$